Problem: Divya and Miguel were asked to find an explicit formula for the sequence $80\,,\,40\,,\,20\,,\,10,...$, where the first term should be $h(1)$. Divya said the formula is $h(n)=80\cdot\left(\dfrac{1}{2}\right)^{{n-1}}$, and Miguel said the formula is $h(n)=160\cdot\left(\dfrac{1}{2}\right)^{{n}}$. Which one of them is right? Choose 1 answer: Choose 1 answer: (Choice A) A Only Divya (Choice B) B Only Miguel (Choice C) C Both Divya and Miguel (Choice D) D Neither Divya nor Miguel
In a geometric sequence, the ratio between successive terms is constant. This means that we can move from any term to the next one by multiplying by a constant value. Let's calculate this ratio over the first few terms: $\dfrac{10}{20}=\dfrac{20}{40}=\dfrac{40}{80}={\dfrac{1}{2}}$ We see that the constant ratio between successive terms is ${\dfrac{1}{2}}$. In other words, we can find any term by starting with the first term and multiplying by ${\dfrac{1}{2}}$ repeatedly until we get to the desired term. Let's look at the first few terms expressed as products: $n$ $1$ $2$ $3$ $4$ $h(n)$ ${80}\cdot\!\left({\dfrac{1}{2}}\right)^{0}$ ${80}\cdot\!\left({\dfrac{1}{2}}\right)^{1}$ ${80}\cdot\!\left({\dfrac{1}{2}}\right)^{2}$ ${80}\cdot\!\left({\dfrac{1}{2}}\right)^{3}$ We can see that every term is the product of the first term, ${80}$, and a power of the constant ratio, ${\dfrac{1}{2}}$. Note that this power is always one less than the term number $n$. This is because the first term is the product of itself and plainly $1$, which is like taking the constant ratio to the zeroth power. Thus, we arrive at the following explicit formula (Note that ${80}$ is the first term and ${\dfrac{1}{2}}$ is the constant ratio): $h(n)={80}\cdot\left({\dfrac{1}{2}}\right)^{{\,n-1}}$ So Divya is definitely right. What about Miguel? We can see that in Miguel's formula, the constant ratio is taken to the $n^{\text{th}}$ power. Let's expand the power in Divya's formula to arrive at a similar expression form: $\begin{aligned} h(n)= &{80}\cdot\left({\dfrac{1}{2}}\right)^{{\,n-1}}\\\\ = & 80\cdot\left(\dfrac{1}{2}\right)^{{n}}\cdot\left(\dfrac{1}{2}\right)^{{-1}}\\\\ = & 80\cdot 2\cdot \left(\dfrac{1}{2}\right)^{{n}} \\\\ = &160\cdot\left(\dfrac{1}{2}\right)^{{n}}\end{aligned}$ We obtained Miguel's formula, which means it's also a correct explicit formula for $h(n)$. Both Divya and Miguel got a correct explicit formula.